series parallel

[pasebay]

i need help finding a missing resistor in a series parallel circuit. My source is 120v. R1 is 12ohms. it is connected to R2, R3 which are parallel to each other. R2 is 20ohms i dont know the current flowing through it. R3 is the missing resistor, but the current flowing through it is 2A. how do i fing out what R3 is?????????????? once i find R3, then i can parallel R2R3, then i can say that they are in series with R1, but!!!!!!!!!! i dont know R3?????????

 

[Klaus]

R3 = 30 Ohm

Since this looks very much like a homework assignment I let you work out how this value was obtained.

Hint: work backwards :wink:

Klaus

 

[audioguru]

Hi Pasebay,
Instead of using a complicated formula, I just used common sense:
1) Quick check that 12 ohms can supply more than 2A. Yeah, way more.
2) If R3 is 20 ohms like R2, then the parallel combination of R2 and R3 is 10 ohms, and the current through that 10 ohms + 12 = 5.45A. Divided by 2 = 2.73A. Too high.
3) Tried 30 ohms for R3. 20 in parallel with 30 = 12 ohms. Therefore the voltage across R3 is 60V and it passes 2A.

I use the same "trial and error" method to calculate a transistor's operating point. R1 and R2 are the voltage divider that feeds the base. The current through R3 is the base current, which is the voltage (minus the B-E drop) divided by the emitter resistor times beta. The base current loads-down the voltage divider. A few trials and I got it.

 

[pasebay]

thanks

i thought of the trial and error method, but i still dont know which formula to use to find the 30ohm resistor???

 

[Phasor]

The "2A" notation is ambiguous. Is that 2A through R3 only, or 2A total battery current?

EDIT: Oops, didn't read your first post properly - ignore my question...

 

[pasebay]

so does anyone know the formula , to find out that R3 equals 30ohms?????

 

[Phasor]

You need to use simultaneous linear equations to solve for I2.

I'm not sure if this is the shortest possible method, but here's how I worked it out:

V2 = 20I2 (from ohms law) <-- this is equation #1

Also, we know that V2 = 120 - V1 , so given that V1 = 12I1 ,
then V2 = 120 - 12I1

I1 is not much good to us, so we need to express I1 in terms of I2 . Therefore I1 = I2 + I3 = I2 + 2
Substituting into above eqn, V2 = 120 - 12(I2 +2), which simplifies to
V2 = 12(8 - I2) <-- this is equation #2

Solve equations #1 and #2 simultaneously, to find that I2 is equal to 3 amps, and V2 = V3 = 60 volts.

Apply ohms law again to R3 --> R3 = 60 / 2 = 30 ohms

 

[checkmate]

It's just simple Kirchoff's Law and Ohm's Law.
Anyway, for more complicated circuits, we usually use the techniques shown in **broken link removed**. In particular, the nodal and mesh methods allow a set of linear equations to be derived for very large systems and solved using Maths software like MATLAB.
OT here, but those are some very power resistors you are working with.

 

[pasebay]

thanks alot guys